Given that $y = \frac{2x^2}{x - 3}$, determine the range of values of $x$ for which $y$ is a decreasing function.

1. Find the first derivative of $y$ with respect to $x$ using the quotient rule.

Let $u = 2x^2$ and $v = x - 3$.
$\frac{du}{dx} = 4x$
$\frac{dv}{dx} = 1$
$\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$
$\frac{dy}{dx} = \frac{(x - 3)(4x) - (2x^2)(1)}{(x - 3)^2}$
$\frac{dy}{dx} = \frac{4x^2 - 12x - 2x^2}{(x - 3)^2}$
$\frac{dy}{dx} = \frac{2x^2 - 12x}{(x - 3)^2}$

2. Set up the inequality for a decreasing function.

For $y$ to be a decreasing function, $\frac{dy}{dx} < 0$.
$\frac{2x^2 - 12x}{(x - 3)^2} < 0$

3. Solve the inequality.

Since $(x - 3)^2 > 0$ for all $x \neq 3$, we must have:
$2x^2 - 12x < 0$
$2x(x - 6) < 0$
The roots are $x = 0$ and $x = 6$.
Thus, $0 < x < 6$.
Since $x = 3$ makes the denominator zero (vertical asymptote), $x \neq 3$.